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\title{\heiti\zihao{2} 习题4.2}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{求下列函数的导函数:}
\subsection{$x \sin x+2 \mathrm{e}^{x}-\sqrt{x}$}
\textbf{解}\quad
$$f^\prime(x)=\sin x+x \cos x+2 e^{x}-\dfrac{1}{2 \sqrt{x}}$$

\subsection{$\sec x+\tan x-\dfrac{1}{\sqrt{x}}$}
\textbf{解}\quad
$$f^{\prime}(x)=\sec  x \tan x+\sec ^{2} x+\dfrac{1}{2 x^{\dfrac{3}{2}}}$$

\subsection{$\mathrm{e}^{x}\left(\tan x-x^{3}+2 \cos x\right)$}
\textbf{解}\quad
$$f^\prime(x)=\left(\tan x-x^{3}+2 \cos x\right) e^{x}+\left(\sec ^{2} x-3 x^{2} -2\sin x\right) e^{x}$$

\subsection{$\left(x^{3}+2 x^{2}-3\right) \ln x$}
\textbf{解}\quad
$$f^{\prime}(x)=\left(3 x^{2}+4 x\right) \ln x+\dfrac{\left(x^{3}+2 x^{2}-3\right)}{x}$$

\subsection{$\left(2^{x}+\log _{3} x\right) \cos x$}
\textbf{解}\quad
$$f^{\prime}(x)=\left(2^{x} \ln 2+\dfrac{1}{x \ln 3}\right) \cos x-\sin x\left(2^{x}+\log _{3} x\right)$$

\subsection{$\dfrac{x^{2} \cos x-\ln x}{\sqrt{x}+3}$}
\textbf{解}\quad
$$f^{\prime}{(x)}=\dfrac{\left(2 x \cos x-x^{2} \sin x-\dfrac{1}{x}\right)(\sqrt{x}+3)-\dfrac{1}{2 \sqrt{x}}\left(x^{2} \cos x-\ln x\right)}{(\sqrt{x}+3)^{2}} $$

\subsection{$\dfrac{3 x-2 \cot x}{\ln x}$}
\textbf{解}\quad
$$f^{\prime}{(x)}=\dfrac{3 x \ln x+2 x \csc ^{2} x \ln x-3 x+2 \cot x}{x \ln ^{2} x}$$

\subsection{$\dfrac{x^{2}+\sec x}{x-\csc x}$}
\textbf{解}\quad
$$f^{\prime} (x)=\dfrac{(2 x+\sec x \tan x)(x-\csc x)-(1+\cot x \csc x)\left(x^{2}+\sec x\right)}{(x-\csc x)^{2}}$$

\section{证明多项式求导后仍是多项式}
\textbf{证}\quad
对于多项式中的任意一个算子$\alpha_ix^i$,对其求导,得$\mathrm{LHS'}=i\alpha_ix^{i-1}$,将其视为$\alpha_{i-1}x^{i-1}$,则其为原函数f(x)得导函数的一个算子.显然由$\alpha_{i-1}x^{i-1}$构成的f(x)的导函数仍是一个多项式.

\section{设函数 $f(x)$ 可导且无零点,证明 : 曲线 $y=f(x)$ 和 $y=f(x) \sin x$ 在相交点处相切.}
\textbf{证}\quad
由于$f(x)无零点$所以其相等则意味着$\sin x=1$即(不妨)$x=\dfrac{\pi}{2}$

$f^\prime\left(\dfrac{\pi}{2}\right)=f^{\prime}\left(\dfrac{\pi}{2}\right) \cdot \sin \dfrac{\pi}{2}+\cos \dfrac{\pi}{2} \cdot f\left(\dfrac{\pi}{2}\right)=f^{\prime}\left(\dfrac{\pi}{2}\right)$

所以其相切

\section{求下列函数的导数:}
\subsection{$\mathrm{e}^{a x} \sin b x$}
\textbf{解}\quad
$$f^{\prime}(x)=a \mathrm{e}^{a x} \sin b x+b \mathrm{e}^{a x} \cos b x$$

\subsection{$\ln [\ln (\ln x)]$}
\textbf{解}\quad
$$f^{\prime}(x)=\dfrac{1}{\ln (\ln x)} \dfrac{1}{\ln x} \dfrac{1}{x}$$

\subsection{$\sqrt{1+x^{2}}$}
\textbf{解}\quad
$$f^{\prime}(x)=\dfrac{x}{\sqrt{1+x^{2}}}$$

\subsection{$\sqrt{1-x^{2}}$}
\textbf{解}\quad
$$f^{\prime}(x)=-\dfrac{x}{\sqrt{1-x^{2}}}$$

\subsection{$\arctan \left(1+x^{2}\right)$}
\textbf{解}\quad
$$f^{\prime}(x)=\dfrac{2 x}{1+\left(1+x^{2}\right)^{2}}$$

\subsection{$\ln (\cos x+\sin x)$}
\textbf{解}\quad
$$f^{\prime}(x)=\dfrac{\cos x-\sin x}{\cos x+\sin x}$$

\subsection{$a^{\sin x}(a>0, a \neq 1)$}
\textbf{解}\quad
$$f^{\prime}(x)=a^{\sin x} \ln a \cdot \cos x$$

\subsection{$\dfrac{\arcsin x}{\sqrt{1-x^{2}}}$}
\textbf{解}\quad
$$f^{\prime}(x)=\dfrac{\dfrac{1}{\sqrt{1-x^{2}}} \sqrt{1-x^{2}}-\arcsin x \dfrac{-x}{\sqrt{1-x^{2}}}}{1-x^{2}}=\dfrac{1+\arcsin x \dfrac{x}{\sqrt{1-x^{2}}}}{1-x^{2}}$$

\subsection{ $x(\cos (\ln x)-\sin (\ln x))$}
\textbf{解}\quad
$$\begin{aligned} f^{\prime}(x) &=[\cos (\ln x)-\sin (\ln x)]+x\left[-\sin (\ln x) \cdot \dfrac{1}{x}-\cos (\ln x) \cdot \dfrac{1}{x}\right] \\ &=-2 \sin (\ln x) \end{aligned}$$


\subsection{$\ln \left(x+\sqrt{a^{2}+x^{2}}\right)$}
\textbf{解}\quad
$$f^{\prime}{(x)}=\dfrac{-2 x^{2}+a^{2}}{\sqrt{a^{2}-x^{2}}}+\dfrac{a^{2}}{\left(a^{2}-x^{2}\right)^{\dfrac{3}{2}}}$$

\subsection{$x \sqrt{a^{2}-x^{2}}+\dfrac{x}{\sqrt{a^{2}-x^{2}}}$}
\textbf{解}\quad
$$f^{\prime}(x)=\sqrt{a^{2}-x^{2}}+\left(-\dfrac{x^{2}}{\sqrt{a^{2}-x^{2}}}\right)+\dfrac{\sqrt{a^{2}-x^{2}}+\dfrac{x^{2}}{\sqrt{a^{2}-x^{2}}}}{\left(\sqrt{a^{2}-x^{2}}\right)^{2}}=\dfrac{-2 x^{2}+a^{2}}{\sqrt{a^{2}-x^{2}}}+\dfrac{a^{2}}{\left(a^{2}-x^{2}\right) \sqrt{a^{2}-x^{2}}}$$

\subsection{$\dfrac{1}{2}\left[x \sqrt{x^{2}-a^{2}}-a^{2} \ln \left(x+\sqrt{x^{2}-a^{2}}\right)\right]$}
\textbf{解}\quad
$$\begin{aligned} f^{\prime}(x) &=\dfrac{1}{2}\left[\sqrt{x^{2}-a^{2}}+\dfrac{x^{2}}{\sqrt{x^{2}-a^{2}}}-a^{2} \dfrac{1+\dfrac{x}{\sqrt{x^{2}-a^{2}}}}{x+\sqrt{x^{2}-a^{2}}}\right.\\ &=\dfrac{1}{2}\left(\dfrac{2 x^{2}-a^{2}}{\sqrt{x^{2}-a^{2}}}-\dfrac{a^{2}}{\sqrt{x^{2}-a^{2}}}\right)=\sqrt{x^{2}-a^{2}} \end{aligned}$$

\subsection{$\mathrm{e}^{a^{x}} \dfrac{a \sin b x-b \cos b x}{\sqrt{a^{2}+b^{2}}}$}
\textbf{解}\quad
$$f^{\prime}(x)=e^{a^{x}} \cdot a^{x} \ln a\left(\dfrac{a \sin b x-b \cos b x}{\sqrt{a^{2}+b^{2}}}\right)+e^{a^{x}}\left(\dfrac{a b \cos b x+b^{2} \sin b x}{\sqrt{a^{2}+b^{2}}}\right)$$

\subsection{$\ln \left[\dfrac{1}{x}+\ln \left(\dfrac{1}{x}+\ln \dfrac{1}{x}\right)\right]$}
\textbf{解}\quad
$$f^{\prime}(x)=\dfrac{x \ln x-1-x-x^{2}}{x(1-x \ln x)\left(1+x \ln \left(\dfrac{1}{x}-\ln x\right)\right)}$$

\section{记 $\sinh x=\dfrac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}, \cosh x=\dfrac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2},$ 分别称为双曲正弦和双曲余弦,证明:}
\subsection{$\cosh ^{2} x-\sinh ^{2} x=1$}
\textbf{证}\quad
$$\cosh ^{2} x-\sinh ^{2} x=\left(\dfrac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}\right)^{2}-\left(\dfrac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}\right)^{2}=\dfrac{\mathrm{e}^{2 x}+\mathrm{e}^{-2 x}+2}{4}-\dfrac{\mathrm{e}^{2 x}+\mathrm{e}^{-2 x}-2}{4}=1$$

\subsection{$(\sinh x)^{\prime}=\cosh x,(\cosh x)^{\prime}=\sinh x$}
\textbf{证}\quad
$(\sinh x)^{\prime}=\left(\dfrac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}\right)^{\prime}=\dfrac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}=\cosh x$

$(\cosh x)^{\prime}=\left(\dfrac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}\right)^{\prime}=\dfrac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}=\sinh x$.

\subsection{求它们的反函数 arcsinh $x, \operatorname{arccosh} x$ 的导数}
\textbf{解}\quad
$\operatorname{arcsinh} x^{\prime}=\dfrac{1}{\cosh y}=\dfrac{1}{\cosh \operatorname{arcsinh} x}$

$\operatorname{arccosh} x^{\prime}=\dfrac{1}{\sinh y}=\dfrac{1}{\operatorname{sinharccosh} x}$

\section{求下列函数的导数:}
\subsection{$\sqrt[x]{x}(x>0)$}
\textbf{解}\quad
设$y=\sqrt[x]{x}(x>0),$ 取对数有 $\ln y=\dfrac{1}{x} \ln x,$ 两边关于 $x$ 求导可得
$$
	\dfrac{y^{\prime}}{y}=-\dfrac{1}{x^{2}} \ln x+\dfrac{1}{x^{2}}=\dfrac{1}{x^{2}}(1-\ln x)
$$
解得$y^{\prime}=y \dfrac{1}{x^{2}}(1-\ln x)=x^{\dfrac{1}{x}-2}(1-\ln x)$

\subsection{$\left(x^{3}+\cos x\right)^{\dfrac{1}{x}}$}
\textbf{解}\quad
设 $y=\sqrt[x]{x^{3}+\cos x},$ 取对数有 $\ln y=\dfrac{1}{x} \ln \left(x^{3}+\cos x\right),$ 两边关于 $x$ 求导可得
$$
	\begin{array}{l}
		\dfrac{y^{\prime}}{y}=-\dfrac{1}{x^{2}} \ln \left(x^{3}+\cos x\right)+\dfrac{1}{x} \dfrac{2 x^{2}-\sin x}{x^{3}+\cos x} ,
		\text { 解得 } y^{\prime}=y\left[-\dfrac{1}{x^{2}} \ln \left(x^{3}+\cos x\right)+\dfrac{1}{x} \dfrac{2 x^{2}-\sin x}{x^{3}+\cos x}\right] \\
		=\left(x^{3}+\cos x\right)^{\dfrac{1}{x}}\left[-\dfrac{1}{x^{2}} \ln \left(x^{3}+\cos x\right)+\dfrac{1}{x} \dfrac{2 x^{2}-\sin x}{x^{3}+\cos x}\right]
	\end{array}
$$

\subsection{$|\cos x|^{x}$}
\textbf{解}\quad
设 $y=|\cos x|^{x},$ 取对数有 $\ln y=x \ln |\cos x|,$ 两边关于 $x$ 求导可得
$$
	\dfrac{y^{\prime}}{y}=\ln |\cos x|+x \dfrac{|\cos x|^{\prime}}{|\cos x|}=\ln |\cos x|-x \tan x
$$
$y^{\prime}=y(\ln |\cos x|-x \tan x)=|\cos x|^{x}(\ln |\cos x|-x \tan x)$

\subsection{$y=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right)$}
\textbf{解}\quad
归纳：

设n=k时,满足LHS‘=$\sum_{i=1}^{k} \prod_{j \neq i}(x-x j)$

则n=k+1时,LHS’=$(\sum_{i=1}^{k} \prod_{j \neq i}(x-x j))\cdot(x-x_{k+1})+\sum_{i=1}^{k} \prod_{j \neq i}(x-x j)=\sum_{i=1}^{k+1} \prod_{j \neq i}(x-x j)$

所以$\mathrm{LHS'}=\sum_{i=1}^{n} \prod_{j \neq i}(x-x j)$

\subsection{$x \dfrac{\sqrt{1-x^{2}}}{\sqrt{1+x^{3}}}$}
\textbf{解}\quad
设 $y=x \dfrac{\sqrt{1-x^{2}}}{\sqrt{1+x^{3}}},$ 当 $x \in(-1,1)$ 时,取对数有 $\ln y=\ln x+\ln \sqrt{1-x^{2}}-\ln \sqrt{1+x^{3}}$
两边关于 $x$ 求导可得
$$
	\begin{aligned}
		\dfrac{y^{\prime}}{y} & =\dfrac{1}{x}+\dfrac{1}{\sqrt{1-x^{2}}} \dfrac{-x}{\sqrt{1-x^{2}}}-\dfrac{1}{\sqrt{1+x^{3}}} \dfrac{3 x^{2}}{2 \sqrt{1+x^{3}}} \\
		                     & =\dfrac{1}{x}-\dfrac{x}{1-x^{2}}-\dfrac{3 x^{2}}{2\left(1+x^{3}\right)}
	\end{aligned}
$$
所以
$$
	\begin{aligned}
		y^{\prime} & =y\left[\dfrac{1}{x}-\dfrac{x}{1-x^{2}}-\dfrac{3 x^{2}}{2\left(1+x^{3}\right)}\right]=x \dfrac{\sqrt{1-x^{2}}}{\sqrt{1+x^{3}}}\left[\dfrac{1}{x}-\dfrac{x}{1-x^{2}}-\dfrac{3 x^{2}}{2\left(1+x^{3}\right)}\right] \\
		           & =x \dfrac{\sqrt{1-x^{2}}}{\sqrt{1+x^{3}}} \dfrac{2\left(1+x^{3}\right)\left(1-x^{2}\right)-2 x^{2}\left(1+x^{3}\right)-3 x^{3}\left(1-x^{2}\right)}{2 x\left(1+x^{3}\right)\left(1-x^{2}\right)}             \\
		           & =\dfrac{2-x^{3}-4 x^{2}-x^{5}}{2\left(1+x^{3}\right)^{\dfrac{3}{2}} \sqrt{1-x^{2}}}=\dfrac{(1+x)^{2}\left(-x^{3}+2 x^{2}-4 x+2\right)}{2(1+x)^{2}\left(1-x+x^{2}\right)^{\dfrac{3}{2}} \sqrt{1-x}}             \\
		           & =\dfrac{-x^{3}+2 x^{2}-4 x+2}{2\left(1-x+x^{2}\right)^{\dfrac{3}{2}} \sqrt{1-x}}
	\end{aligned}
$$

\subsection{$x^{\sin x}+e^{x^{x}}$}
\textbf{解}\quad
$$f^\prime(x)=x^{\sin x} \cos x \ln x+x^{\sin x-1} \sin x+\mathrm{e}^{x^{x}} \cdot x^{x}(\ln x+1)$$

\section{设 $f(x)$ 是 $n$ 次多项式,它有 $k$ 个相异实根 $x_{1}, x_{2}, \cdots, x_{k},$ 对应重数为 $n_{1}, n_{2}, \cdots, n_{k},$ 且有 $n_{1}+n_{2}+\cdots+n_{k}=n .$ 证明 :
$$
	f^{\prime}(x)=f(x)\left(\sum_{i=1}^{k} \dfrac{n_{i}}{x-x_{i}}\right)
$$}
\textbf{证}\quad
$f(x)$ 是 $n$ 次多项式,并且恰好有 $n$ 个实根,所以可以将 $f(x)$ 分解为
$$
	f(x)=a\left(x-x_{1}\right)^{n_{1}}\left(x-x_{2}\right)^{n_{2}} \cdots\left(x-x_{k}\right)^{n_{k}}
$$
则由导数的线性性质,有
$$
	\begin{aligned}
		f^{\prime}(x) & =\left[a\left(x-x_{1}\right)^{n_{1}}\left(x-x_{2}\right)^{n_{2}} \cdots\left(x-x_{k}\right)^{n_{k}}\right]^{\prime}                                             \\
		              & =\sum_{i=1}^{k}\left[n_{i}\left(x-x_{i}\right)^{n_{i}-1} \dfrac{f(x)}{\left(x-x_{i}\right)^{n_{i}}}\right]=f(x) \sum_{i=1}^{k}\left(\dfrac{n_{i}}{x-x_{i}}\right)
	\end{aligned}
$$

\section{已知 $f(x), g(x)$ 为 $x$ 的可导函数,求下列函数 (在可导点处)的导数:}
\subsection{$\sqrt{f^{2}(x)+g^{2}(x)}$}
\textbf{解}\quad
$$\left[\sqrt{f^{2}(x)+g^{2}(x)}\right]^{\prime}=\dfrac{f(x) f^{\prime}(x)+g(x) g^{\prime}(x)}{\sqrt{f^{2}(x)+g^{2}(x)}}$$

\subsection{$\tan \dfrac{f(x)}{g(x)}$}
\textbf{解}\quad
$$\quad\left[\tan \dfrac{f(x)}{g(x)}\right]^{\prime}=\sec ^{2}\left[\dfrac{f(x)}{g(x)}\right] \cdot\left[\dfrac{f(x)}{g(x)}\right]^{\prime}=\dfrac{f^{\prime}(x) g(x)-f(x) g^{\prime}(x)}{g^{2}(x)} \sec ^{2}\left[\dfrac{f(x)}{g(x)}\right]$$

\subsection{$\log _{f(x)} g(x)$}
\textbf{解}\quad
$$
	\begin{aligned}
		 & y=\log _{f(x)} g(x)= \dfrac{\ln g(x)}{\ln f(x)}, \text { 则 } \\&
		\begin{aligned}
			y^{\prime} & =\dfrac{\dfrac{g^{\prime}(x)}{g(x)} \ln f(x)-\dfrac{f^{\prime}(x)}{f(x)} \ln g(x)}{\ln ^{2} f(x)}
			=\dfrac{f(x) g^{\prime}(x) \ln f(x)-g(x) f^{\prime}(x) \ln g(x)}{g(x) f(x) \ln f(x)}
		\end{aligned}
	\end{aligned}
$$

\subsection{$f\left([g(x)]^{2}\right)$}
\textbf{解}\quad
$$y^{\prime}=2 f^{\prime}\left(g^{2}(x)\right) \cdot g(x) g^{\prime}(x)$$

\section{设 $f(x)$ 在 $x=x_{0}$ 处可导, $g(x)$ 在 $x=x_{0}$ 处不可导,证明 $f(x)+g(x)$ 在 $x=x_{0}$ 处不可导}
\textbf{反证}\quad
假设$f(x)+g(x)在x_{0}$处可导,则$g(x)=f(x)+g(x)-g(x)$在$x_{0}$处也可导,矛盾.

\section{在 $\mu$ 满足什么条件下函数 $f(x)=\left\{\begin{array}{ll}|x|^{\mu} \sin \dfrac{1}{x}, & \text { 当 } x \neq 0 \text { 时 } \\ 0, & \text { 当 } x=0 \text { 时 }\end{array}\right.$}
\subsection{在 $x=0$ 处连续}
\textbf{解}\quad
满足$\lim _{x \rightarrow 0}|x|^{\mu} \sin \dfrac{1}{x}=0$必有$\mu>0$,当$\mu<0$时极限不存在,$\mu=0$时$\sin{\dfrac{1}{x}}$在$0$处波动

\subsection{在 $x=0$ 处可导}
\textbf{解}\quad
$f_{+}^{\prime}(0)=\lim _{x \rightarrow 0^{+}} \dfrac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \dfrac{|x|^{\mu} \sin \dfrac{1}{x}-0}{x}=\lim _{x \rightarrow 0^{+}} x^{\mu-1} \sin \dfrac{1}{x}$
$$
	f_{-}^{\prime}(0)=\lim _{x \rightarrow 0^{-}} \dfrac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{-}} \dfrac{(-x)^{\mu} \sin \dfrac{1}{x}-0}{x}=-\lim _{x \rightarrow 0^{-}} x^{\mu-1} \sin \dfrac{1}{x}
$$
当 $\mu>1$ 时,上述极 限为无穷小量乘以有界变量, 所以 $f^{\prime}(0)=0$; 当 $1>\mu>0$ 时,上述极限 为无穷大量乘以有界变量,所以 $f^{\prime}(0)$ 不存在,故当 $\mu>1$ 时, $f(x)$ 是可导函数.当$\mu=0$时波动.

\subsection{在 $x=0$ 处其导函数连续.}
\textbf{解}\quad
当$\mu>1$时函数可导.

$f^{\prime}(x)=\left\{\begin{array}{ll}0, & x=0 \\ \mu x^{\mu-1} \sin \dfrac{1}{x}-x^{\mu-2} \cos \dfrac{1}{x}, & x>0 \\ -\left[\mu(-x)^{\mu-1} \sin \dfrac{1}{x}+(-x)^{\mu-2} \cos \dfrac{1}{x}\right], & x<0\end{array}\right.$

同上理,$x\rightarrow0$时$f^\prime(x) \rightarrow0$所以必须要求$\mu>2$

\section{在 $\mu$ 满足什么条件下函数 $f(x)=\left\{\begin{array}{ll}|x|^{\mu} \arctan \dfrac{1}{x}, & \text { 当 } x \neq 0 \text { 时 } \\ 0, & \text { 当 } x=0 \text { 时 }\end{array}\right.$}
\subsection{在 $x=0$ 处连续 }
\textbf{解}\quad
$$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}|x|^{\mu} (\dfrac{\pi}{2})=\lim _{x \rightarrow 0^{+}} x^{\mu} (\dfrac{\pi}{2})\\
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}|x|^{\mu} \arctan \dfrac{1}{x}=\lim _{x \rightarrow 0^{-}}(-x)^{\mu} (-\dfrac{\pi}{2})$$

左右极限相等需要$\mu>0$

\subsection{在 $x=0$ 处可导}
\textbf{解}\quad
$f_{+}^{\prime}(0)=\lim _{x \rightarrow 0^{+}} \dfrac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \dfrac{|x|^{\mu} \arctan \dfrac{1}{x}-0}{x}=\lim _{x \rightarrow 0^{+}} x^{\mu-1} (\dfrac{\pi}{2})$
$$
f_{-}^{\prime}(0)=\lim _{x \rightarrow 0^{-}} \dfrac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{-}} \dfrac{(-x)^{\lambda} \arctan \dfrac{1}{x}-0}{x}=-\lim _{x \rightarrow 0^{-}} x^{\lambda-1} (-\dfrac{\pi}{2})
$$

当 $\mu>1$ 时,上述极 限为无穷小量乘以有界变量, 所以 $f^{\prime}(0)=0 ;$ 当 $1>\mu>0$ 时,上述极限 为无穷大量乘以有界变量,所以 $f^{\prime}$ (0)不存在,故当 $\mu>1$ 时, $f(x)$ 是可导函数.

\subsection{在 $x=0$ 处其导函数连续}
\textbf{解}\quad
当 $\mu>1$ 时,是可导函数, 由 (2) 可 知 $f^{\prime}(0)=0 .$

当 $x>0$ 时, $\begin{aligned} f^{\prime}(x) &=\left(x^{\mu} \arctan \dfrac{1}{x}\right)^{\prime}=\mu x^{\mu-1} \arctan \dfrac{1}{x}+x^{\mu} \dfrac{1}{1+\dfrac{1}{x^{2}}}\left(\dfrac{1}{x}\right)^{\prime} \\ &=\mu x^{\mu-1} \arctan \dfrac{1}{x}+x^{\mu} \dfrac{x^{2}}{1+x^{2}} \dfrac{-1}{x^{2}}=\mu x^{\mu-1} \arctan \dfrac{1}{x}-x^{\mu} \dfrac{1}{1+x^{2}} \end{aligned}$

当 $x<0$ 时,
$$
\begin{aligned}
f^{\prime}(x) &=\left[(-x)^{\mu} \arctan \dfrac{1}{x}\right]^{\prime}=-\mu(-x)^{\mu-1} \arctan \dfrac{1}{x}+(-x)^{\mu} \dfrac{1}{1+\dfrac{1}{x^{2}}}\left(\dfrac{1}{x}\right)^{\prime} \\
&=-\mu(-x)^{\mu-1} \sin \dfrac{1}{x}+(-x)^{\mu} \dfrac{x^{2}}{1+x^{2}} \dfrac{-1}{x^{2}} \\
&=-\left[\mu(-x)^{\mu-1} \arctan \dfrac{1}{x}+(-x)^{2} \dfrac{1}{1+x^{2}}\right]
\end{aligned}
$$

而显然当$\mu>1$时即可满足左右导函数在0处的极限都是0,若$\mu<1$则导函数在0点极限不存在.若$\mu=1$,则在零点极限不为0,所以也不满足导函数连续的条件.

综上,$\mu>1$

\end{document}